Afugitive tries to hop on a freight train traveling at a constant speed of 4.5 m/s. just as an empty box car passes him, the fugitive starts from rest and accelerates at a = 3.4 m/s2 to his maximum speed of 8.0 m/s. (a) how long does it take him to catch up to the empty box car? . s. (b) what is the distance traveled to reach the box car? .

Ok, 
Let's check to see if he catches it before he reaches 8 m/s. 
Velocity = acceleration * time, V = at, 
so 8 m/s = 3.4 m/s * t, so t = 2.35 sec 
in that time he moved 
Distance = Initial Velocity * time + 1/2 acceleration * time * time 
S = 0*2.35 + 1/2*3.4*2.35*2.35 = 9.4 meters 
in 2.35 seconds, the boxcar moved 
S = V*t = 4.5 * 2.35 = 10.59 meters, so he still needs to catch up 
10.59 m - 9.4 m = 1.19 meters. Since he is now moving at 8 m/s and the train is moving at 4.5 m/s, his closing speed is (8 - 4.5) = 3.5 m/s 

It will take him 1.19m/3.5m/s = .34 seconds more to catch up. 

So answer A is 2.35 sec + .34 sec = 2.69 seconds. 
For answer B we use the speed of the train since it was constant, 
Distance = Velocity * time = 4.5 m/s * 2.69 s = 12.1 meters.


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